\(\int \frac {-1+x^2}{x^2 (1+x^2) \sqrt [4]{x^2+x^4}} \, dx\) [391]

Optimal result

Integrand size = 27, antiderivative size = 32 \[ \int \frac {-1+x^2}{x^2 \left (1+x^2\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {2 \left (1+7 x^2\right ) \left (x^2+x^4\right )^{3/4}}{3 x^3 \left (1+x^2\right )} \]

[Out]

2/3*(7*x^2+1)*(x^4+x^2)^(3/4)/x^3/(x^2+1)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2081, 464, 270} \[ \int \frac {-1+x^2}{x^2 \left (1+x^2\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {14 x}{3 \sqrt [4]{x^4+x^2}}+\frac {2}{3 \sqrt [4]{x^4+x^2} x} \]

[In]

Int[(-1 + x^2)/(x^2*(1 + x^2)*(x^2 + x^4)^(1/4)),x]

[Out]

2/(3*x*(x^2 + x^4)^(1/4)) + (14*x)/(3*(x^2 + x^4)^(1/4))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt [4]{1+x^2}\right ) \int \frac {-1+x^2}{x^{5/2} \left (1+x^2\right )^{5/4}} \, dx}{\sqrt [4]{x^2+x^4}} \\ & = \frac {2}{3 x \sqrt [4]{x^2+x^4}}+\frac {\left (7 \sqrt {x} \sqrt [4]{1+x^2}\right ) \int \frac {1}{\sqrt {x} \left (1+x^2\right )^{5/4}} \, dx}{3 \sqrt [4]{x^2+x^4}} \\ & = \frac {2}{3 x \sqrt [4]{x^2+x^4}}+\frac {14 x}{3 \sqrt [4]{x^2+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {-1+x^2}{x^2 \left (1+x^2\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {2 \left (1+7 x^2\right )}{3 x \sqrt [4]{x^2+x^4}} \]

[In]

Integrate[(-1 + x^2)/(x^2*(1 + x^2)*(x^2 + x^4)^(1/4)),x]

[Out]

(2*(1 + 7*x^2))/(3*x*(x^2 + x^4)^(1/4))

Maple [A] (verified)

Time = 0.99 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69

method	result	size
gosper	\(\frac {\frac {14 x^{2}}{3}+\frac {2}{3}}{x \left (x^{4}+x^{2}\right )^{\frac {1}{4}}}\)	\(22\)
risch	\(\frac {\frac {14 x^{2}}{3}+\frac {2}{3}}{x \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\)	\(24\)
pseudoelliptic	\(\frac {\frac {14 x^{2}}{3}+\frac {2}{3}}{x \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\)	\(24\)
trager	\(\frac {2 \left (7 x^{2}+1\right ) \left (x^{4}+x^{2}\right )^{\frac {3}{4}}}{3 x^{3} \left (x^{2}+1\right )}\)	\(29\)
meijerg	\(\frac {\frac {8 x^{2}}{3}+\frac {2}{3}}{x^{\frac {3}{2}} \left (x^{2}+1\right )^{\frac {1}{4}}}+\frac {2 \sqrt {x}}{\left (x^{2}+1\right )^{\frac {1}{4}}}\)	\(33\)

[In]

int((x^2-1)/x^2/(x^2+1)/(x^4+x^2)^(1/4),x,method=_RETURNVERBOSE)

[Out]

2/3*(7*x^2+1)/(x^4+x^2)^(1/4)/x

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {-1+x^2}{x^2 \left (1+x^2\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {2 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}} {\left (7 \, x^{2} + 1\right )}}{3 \, {\left (x^{5} + x^{3}\right )}} \]

[In]

integrate((x^2-1)/x^2/(x^2+1)/(x^4+x^2)^(1/4),x, algorithm="fricas")

[Out]

2/3*(x^4 + x^2)^(3/4)*(7*x^2 + 1)/(x^5 + x^3)

Sympy [F]

\[ \int \frac {-1+x^2}{x^2 \left (1+x^2\right ) \sqrt [4]{x^2+x^4}} \, dx=\int \frac {\left (x - 1\right ) \left (x + 1\right )}{x^{2} \sqrt [4]{x^{2} \left (x^{2} + 1\right )} \left (x^{2} + 1\right )}\, dx \]

[In]

integrate((x**2-1)/x**2/(x**2+1)/(x**4+x**2)**(1/4),x)

[Out]

Integral((x - 1)*(x + 1)/(x**2*(x**2*(x**2 + 1))**(1/4)*(x**2 + 1)), x)

Maxima [F]

\[ \int \frac {-1+x^2}{x^2 \left (1+x^2\right ) \sqrt [4]{x^2+x^4}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} {\left (x^{2} + 1\right )} x^{2}} \,d x } \]

[In]

integrate((x^2-1)/x^2/(x^2+1)/(x^4+x^2)^(1/4),x, algorithm="maxima")

[Out]

-2/21*(8*x^5 + 7*(x^3 + x)*x^2 + 9*x^3 + x)/((x^(9/2) + x^(5/2))*(x^2 + 1)^(1/4)) + integrate(8/21*(4*x^4 + x^
2 - 3)/((x^(13/2) + 2*x^(9/2) + x^(5/2))*(x^2 + 1)^(1/4)), x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.59 \[ \int \frac {-1+x^2}{x^2 \left (1+x^2\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {2}{3} \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {3}{4}} + \frac {4}{{\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}} \]

[In]

integrate((x^2-1)/x^2/(x^2+1)/(x^4+x^2)^(1/4),x, algorithm="giac")

[Out]

2/3*(1/x^2 + 1)^(3/4) + 4/(1/x^2 + 1)^(1/4)

Mupad [B] (verification not implemented)

Time = 4.95 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {-1+x^2}{x^2 \left (1+x^2\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {2\,{\left (x^4+x^2\right )}^{3/4}\,\left (7\,x^2+1\right )}{3\,x^3\,\left (x^2+1\right )} \]

[In]

int((x^2 - 1)/(x^2*(x^2 + x^4)^(1/4)*(x^2 + 1)),x)

[Out]

(2*(x^2 + x^4)^(3/4)*(7*x^2 + 1))/(3*x^3*(x^2 + 1))